The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is $2.0\mathrm{~g~cm}^{-3}$. The ratio of the molecular weights of the solute and solvent, $\left(\frac{MW_{\text{solute}}}{MW_{\text{solvent}}}\right)$, is

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Solutions

The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is

2.0 g cm^{- 3}

. The ratio of the molecular weights of the solute and solvent,

(\frac{M W _{solute}}{M W _{solvent}})

, is

Answer: 9

Solution

Let

n_{A} = 0.9

mol (solvent) and

n_{B} = 0.1

mol (solute). Molality

m = \frac{n _{B} \times 1000}{n _{A} \times M _{A}} = \frac{0.1 \times 1000}{0.9 M _{A}} = \frac{1000}{9 M _{A}}

. Molarity

M = \frac{n _{B} \times 1000 \times ρ}{n _{A} M _{A} + n _{B} M _{B}} = \frac{0.1 \times 1000 \times 2.0}{0.9 M _{A} + 0.1 M _{B}} = \frac{200}{0.9 M _{A} + 0.1 M _{B}}

. Setting

m = M

\frac{1000}{9 M _{A}} = \frac{200}{0.9 M _{A} + 0.1 M _{B}}

. Cross-multiplying:

1000 (0.9 M_{A} + 0.1 M_{B}) = 1800 M_{A}

, so

900 M_{A} + 100 M_{B} = 1800 M_{A}

, giving

100 M_{B} = 900 M_{A}

, thus

M_{B} / M_{A} = 9

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Integer answer

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JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Solutions

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